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A Defining the surface gravity of a black hole

  1. Apr 20, 2017 #1
    Consider the following two definitions of the surface gravity of a black hole (taken from page 23 of Thomas Hartman's lecture notes(http://www.hartmanhep.net/topics2015/) on Quantum Gravity):

    1. The surface gravity is the acceleration due to gravity near the horizon (which goes to infinity) times the redshift factor (which goes to zero).

    2. If you stand far away from the black hole holding a fishing pole, and dangle an object on your fishing line near so it hovers near the horizon, then you will measure the tension in your fishing line to be ##\kappa M_{\text{object}}##.


    What does it mean for the acceleration due to gravity near the horizon to go to infinity?

    Why do we multiply by the redshift factor and what does it mean for the redshift factor to go to zero?
  2. jcsd
  3. Apr 20, 2017 #2


    Staff: Mentor

    It means that the proper acceleration of an observer hovering at a constant altitude above the hole's horizon increases without bound as the altitude goes to zero.

    To obtain the force that must be exerted by an observer at a much, much higher altitude (in the idealized case under consideration, at "infinity") on a rope, or fishing line, or whatever, that then transmits the force to the object that is being held at a constant altitude very close to the horizon. This force is "redshifted" compared to the force felt by the object itself.

    Given an observer holding the fishing pole at infinity, the redshift factor between that observer and an object at the bottom of the fishing line, at some altitude above the horizon, goes to zero as the altitude goes to zero.

    Mathematically, the two expressions in question are the proper acceleration ##a## felt by the object hovering close to the horizon and the redshift factor ##\chi## between that object and the observer at infinity, given by:

    a = \frac{M}{r^2} \frac{1}{\sqrt{1 - 2M / r}}

    \chi = \sqrt{1 - \frac{2M}{r}}

    The force (per unit mass) exerted by the observer at infinity on the fishing line is then ##a \chi##, and the factor of ##\sqrt{1 - 2M / r}## obviously cancels in this product, so that force is finite as ##r \rightarrow 2M## (i.e., as the altitude above the horizon goes to zero), even though ##a \rightarrow \infty## and ##\chi \rightarrow 0## in this limit.
  4. Apr 20, 2017 #3
    Thank you for your beautiful explanation.

    What does it mean to redshift a force? I only understand the redshift of wavelengths (due to a curved space-time or accelerating space-time) as the stretching of waves before they reach the observer. I am not sure how to import this idea to explain the redshift of forces.

    Some other questions:

    You mention that the proper acceleration felt by the object hovering close to the horizon goes to infinity. Is this merely an artifact of Schwarzchild coordinates?

    I fail to understand why the force has to be product of ##a## with ##\chi##. It's obvious that in the limit ##r \to \infty##, the force is ##M/r^{2}##. This is Newtonian gravity and holds in flat spacetime (Schwarzchild spacetime is asympototically flat). But I fail to understand why the effect of curvature is to modify the force by a factor of the redshift, where this redshift tries to dilute the force as the object gets closer to the horizon.
  5. Apr 20, 2017 #4


    Staff: Mentor

    It means that, because of the geometry of spacetime, the force that needs to be exerted to hold an object static decreases as the altitude of the point where the force is exerted (as compared to the altitude where the object is--obviously this requires a rope, fishing line, or something else to transmit the force) increases. The term "redshift" is used by analogy with the more familiar case of light redshifting as it climbs in a gravitational field.

    No. It's an invariant.

    Remember that we are talking about the force (per unit mass) exerted by the distant observer, who is holding the upper end of the fishing line. The force felt by the object, at the lower end of the fishing line, is not redshifted; it is just ##a##.
  6. Apr 20, 2017 #5
    I understand intuitively that the force exerted by me to hold you steady near the horizon is a function of my distance from you. (This is due to the curvature of spacetime.) What I fail to understand if why that force must decrease as I move further and further away from you. Surely, it must increase as the spacetime wants you to pass through the horizon and sink into the black hole, and I am trying my level best to keep you steady against that force.

    Would I have to grind through the calculation to understand this in detail? Is there no intuitive explanation?

    In this case, does it make sense to talk about the proper acceleration of the person when he hits the horizon? With what proper acceleration is he accelerating as he hits the horizon?
  7. Apr 20, 2017 #6


    Staff: Mentor

    That's not quite what we're talking about. Let's take a step back and describe the scenario in more detail.

    We are at infinity holding a fishing pole to which a line is attached. Hanging from the line, at some finite altitude, is an object. Call the radial coordinate of the object ##r##. Then we have two questions:

    (1) What force do we have to exert on the fishing line, at infinity, to hold the object static, as ##r \rightarrow 2M##?

    (2) What force does the object, at the bottom of the fishing line at radius ##r##, experience as ##r \rightarrow 2M##?

    The answers, mathematically are: (1) We have to exert a force ##a \chi = M / r^2##. (2) The object experiences a force ##a = M / \left( r^2 \sqrt{1 - 2M / r} \right)##.

    Note carefully that both of these forces change as ##r## changes. But they change differently.

    That's not a good way to think of it; the "amount of spacetime" between the observer and the object is not the key factor.

    To understand, heuristically, why the force redshifts, we have to delve deeper into what "force" is. The general definition is ##F = dp / d\tau##, where ##p## is momentum and ##\tau## is proper time. Here "momentum" has to be interpreted carefully, because intuitively the object is static so its momentum isn't changing. But the object is static only because it is held static by a force; if it were not held, it would free-fall downward. So "momentum" here really means something like "the momentum we have to prevent being added to the object by holding it static instead of letting it free-fall".

    With that out of the way, the key is the proper time ##\tau##. Since the object is deep in a gravity well and being held static, it experiences a large amount of gravitational time dilation, relative to the observer at infinity. The force required to hold the object static, at altitude ##r##, is ##dp / d\tau## with ##\tau## being the object's proper time. But the force required at infinity, to keep enough tension in the fishing line to hold the object static, is ##dp / d\tau## with ##\tau## being the observer's proper time, at infinity instead of radius ##r##. And since the observer's proper time is running a lot faster, relative to the object, it can exert much less force, since its ##d \tau## is in the denominator.

    This is not a fully rigorous argument (that would require quite a bit more math), but it should give a good idea of why the force redshifts.
  8. Apr 20, 2017 #7


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    The most direct argument, I think, is to consider seriously the physical characteristics of a massless string. The string has some stress-energy tensor ##T^{ab}##. It's easiest if we use coordinates where the string has a constant cross section. The density terms will be zero, because the string is massless - it will only have tension terms, that support the weight in question, that act along the direction of the string. You can call the single non-zero term ##T^{\hat{1}\hat{1}}##, and with a constant cross section, the tension will be constant across the cross-section of the string, and constant with time.

    The divergence of the stress energy tensor is zero. ##\nabla_a T^{ab} = 0##. This is a pain to work out, though. (I think I did it once, I don't even recall for sure if I did or if I gave up.)

    The less direct argument, due to Wald (exercise 4 in chapter 6) uses a conservation of energy argument. I'm not sure how enlightening the exercise will be - since it only gives hints - but it's a source you can look up if you're ambitious. Wald, "General Relativity", pg 158, problem 4 in chapter 6.

    The sloppy approach is to blame it all on time dilation. Recall that ordinary force is the rate of change of momentum with respect to time, and that time is, again being sloppy, running at different rates at infinity and for the observer hovering over the event horizon.

    No. Proper acceleration isn't coordinate dependent, so it's not an artifact of the coordinate. Only a massless particle can remain stationary at the event horizon - a small pulse of light forever moving outward. The infinite proper acceleration is just a mathematical consequence of the fact that no massive particle can be stationary at the event horizon.
  9. Apr 20, 2017 #8


    Staff: Mentor

    Which is the heuristic argument I gave. I prefer "heuristic" to "sloppy". :wink:

    The reason the heuristic argument works, though, is that it is based on a key factor that comes into play when calculating the covariant divergence of the stress-energy tensor along the string and thereby deriving an equation for the tension in the string as a function of radial coordinate.
  10. Apr 20, 2017 #9
    So, you're saying that a massive object at the event horizon has an infinite acceleration. I understand that the acceleration of the massive object at the event horizon must be a number which cannot be compensated for with an acceleration in the opposite direction. This would allow the massive object to escape even after reaching the event horizon and this is not allowed - this I understand.

    But infinite accelerations are not physical!!!!! How can the massive object then have an infinite acceleration at the event horizon?
  11. Apr 20, 2017 #10
    You guys make me want to quit. How come you know so much physics? Should I be reading textbooks on general relativity if I would like to become as knowledgable as you guys or does your kind of knowledge only comes with years and years of practice in the field? I feel lost!!! :sorry:
  12. Apr 20, 2017 #11


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    Not exactly. I'm saying that if a massive object were stationary at the event horizon, it would have an infinite acceleration. But the real point is - a massive object can't be stationary at the event horizon. Only a massless object (like light) can do that.

    A massive object free-falling through the event horizon would have a proper acceleration of zero - this is the definition of free fall. A rocket could have any finite proper acceleration its engines could deliver at the event horizon - but it would fall into the black hole, it could never have enough thrust to hold station.
  13. Apr 20, 2017 #12
    Ah! I see! So, this is exactly why you, at infinity, can't pull a massive object out from the event horizon (and beyond) once it reaches the event horizon.
  14. Apr 21, 2017 #13


    Staff: Mentor

    No. I have said nothing whatever about the proper acceleration of an object at the horizon. At the horizon, it is impossible for a massive object to be static; it must be falling inwards, as pervect said (and as he said, this is true regardless of how large the object's proper acceleration is). So nothing we have discussed applies to massive objects at the horizon, since we have only been discussing massive objects that are static.
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